# Trignometric equation, proof

1. Feb 16, 2015

### Suraj M

1. The problem statement, all variables and given/known data

$$\cos(2x) × \cos(\frac x2) - \cos(3x)× \cos (\frac{9}{2} x) = \sin(5x)× \sin \frac {5}{2}x$$

2. Relevant equations
$$\cos(x) + \cos(y) = 2 \cos{\frac{x+y}{2}} \cos{\frac{x-y}{2}}$$
$$\cos(x) - \cos (y) = -2 \sin \frac{x+y}{2} \sin \frac{x-y}{2}$$

3. The attempt at a solution
I don't know how to start, as there are no common angles(2x,x/2,3x...).
How do i start off?

Last edited: Feb 16, 2015
2. Feb 16, 2015

### Orodruin

Staff Emeritus
Why do you not simply start expanding your products using the equations you have quoted and see what you get?

3. Feb 16, 2015

### vela

Staff Emeritus
First, start by using parentheses when necessary. It's not true that
$$\cos x + \cos y = 2\cos\left(x+\frac y2\right)\cos\left(x-\frac y2\right)$$ as you wrote.

It might help to write the identities using variables other than x and y to avoid confusion, e.g.
$$\cos a + \cos b = 2\cos \frac{a+b}2 \cos \frac {a-b}2.$$ To deal with the term $\cos 2x \cos \frac x2$, you want $\frac{a+b}2 = 2x$ and $\frac{a-b}2 = \frac x2$. Solve for $a$ and $b$. What do you get?

4. Feb 16, 2015

### Suraj M

I actually don't know how you right those equations using parentheses, so pardon this:
$a = {\frac{5}{2}} x$ and $b = {\frac{3}{2}} x$
and then if i do the same for the second term... $c = {\frac{15}{2}} x$ and $d = {\frac{-3}{2}} x$
so then my LHS becomes:
$${\frac{1}{2}}\left({\cos \left({\frac{5}{2}} x \right) +\cos{\frac{3}{2}}x } \right)- {\frac{1}{2}} \left( [\cos {\frac{15}{2}} x + \cos{\frac{3}{2}} x ] \right)$$
$$= {\frac{1}{2}}\left[(-2) × \sin(-5x) × \sin{\frac{5}{2}}x \right]$$
= RHS
, Thanks a lot
now could you please tell me how to right those equations using parentheses,please

Last edited: Feb 16, 2015
5. Feb 16, 2015

### PWiz

6. Feb 16, 2015

### vela

Staff Emeritus
cos ((x+y)/2) instead of cos (x+y/2)